Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$.

Explanation: Let the two disjoint subsets be $A$ and $B$, and let $C = S-(A+B)$. For each $i \in S$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$.
However, there are $2^{10}$ ways to organize the elements of $S$ such that $A = \emptyset$ and $S = B+C$, and there are $2^{10}$ ways to organize the elements of $S$ such that $B = \emptyset$ and $S = A+C$. But, the combination such that $A = B = \emptyset$ and $S = C$ is counted twice.
Thus, there are $3^{10}-2\cdot2^{10}+1$ ordered pairs of sets $(A,B)$. But since the question asks for the number of unordered sets $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.